Question

The sum of four consecutive terms which are in AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15 find the number​

Answer 1

Answer

Let a be the first term

Let d be the common difference

Then, Let the four consecutive terms be (a-3d), (a-d), (a+d), (a+3d)

A/Q we get,

$a-3d + a-d + a+d +a-3d = 32$

-d and +d , -3d and +3d get cancelled

So, $4a = 32$

$\bold{a = 8}$

$\frac{(a-3d)(a+3d)}{(a-d)(a+d)}$ = $\frac{7}{15}$

$\frac{a^2- (3d)^2}{a^2- d^2)}$ = $\frac{7}{15}$

By cross multiplying,

$15(a^2 - 9d^2) = 7(a^2 - d^2)$

$15a^2 - 135d^2 = 7a^2 - 7d^2$

$15a^2 - 7a^2 = 135d^2 - 7d^2$

$8a^2 = 128d^2$

As we got the value of a=8

putting in the equations,

$8(8)^2 = 128d^2$

$128d^2 = 512$

$d^2 = \frac{512}{128}$

$d^2 = 4$

$\bold{d = 2}$

Putting the values in our assumed terms

we get,   2, 6, 10, 14

Answer 2

Answer:

 2 , 6, 10 , 14 .

Step-by-step explanation:

Let four consecutive number as

a - 3 d , a - d , a + d , a + 3 d .

Given their sum is 32 .

a + a + a + a + 3 d - 3 d = 32

4 a = 32

a = 8 .

Now :

Also given the ratio of the product of the first and the last term to product of two middle term to the product of two middle terms is 7:15 .

$\displaystyle{\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)} = \dfrac{7}{15} }\\\\\\\displaystyle{\dfrac{a^2-(3d)^2}{a^2-d^2} = \dfrac{7}{15} }$

We have a = 8

$\displaystyle{\dfrac{64-(3d)^2}{64-d^2} = \dfrac{7}{15} }\\\\\\\displaystyle{\dfrac{64-9d^2}{64-d^2} = \dfrac{7}{15} }$

960 - 135 d² = 448 - 7 d²

135 d² - 7 d²  = 960 - 448

128 d²  = 512

d²  = 4

d = ± 2

When d = 2

Numbers are ,

a - 3 d , a - d ,  a + d , a + 2 d

= > 8 - 6 = 2

= > 8 - 2 = 6

= > 8 + 2 = 10

= > 8 + 6 = 14

When d = - 2

= > 8 + 6 = 14

= > 8 + 2 = 10

= > 8 - 2 = 6

= > 8 - 6 = 2 .

But both ways we get same numbers.

i.e.  2 , 6, 10 , 14 .